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I just started mechanics of materials, and it seems that my knowledge from statics has gotten rusty over the summer. But there was a new concept introduced in the first example problem today.

There was a horizontal bar connected to a wall (fixed support). I understand that with a fixed support, there are horizontal and vertical reaction forces and a moment as well. However, when the professor cut the horizontal bar to calculate the internal loadings, he drew vectors for vertical and horizontal forces AND a moment. Since I didn't use this approach in statics, I don't get the reasoning behind adding a vector for the moment when the body is cross sectioned. Can someone explain to me why a moment is added at the newly cut end?

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To put it simply your loads are causing the beam/bar to bend right? So at the cut we need to consider the internal bending moment ($M$).

Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three internal forces you need to consider:

  1. Shear Force ($V$ or $Q$) perpendicular to the longitudinal axis of the member.

  2. Axial Force ($N$) along the longitudinal axis of the member.

  3. Bending Moment ($M$) about the z-axis which is coming out of the plane/page.

We consider these three internal loads because we know that they can be experienced by the member along its length due to the reactions at the supports and the external loads. The one you seem confused about is the moment ($M$). Nearly all the time in statics your loads on the beam will cause bending and this why we need to add a force vector for bending moment whenever we take a cut. Imagine taking a ruler, holding it at both ends and causing it to bend in the middle - a bending force is there that needs to be considered.

Now just because we've taken a cut and considered that these three internal loads, it doesn't mean they ALWAYS exist. They MIGHT exist. For instance, if you take a cut of the beam/bar where shear force is 0, then bending moment is typically at its maximum. At the point where no internal shear force exists (i.e. it is switching direction), bending moment happens to be maximum. Another example is at free ends the bending moment is typically zero.

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  • $\begingroup$ Thank you so much for the clear explanation. So I must CONSIDER all three types including the moment because the external forces on the body can cause bending. If I may ask another question just to clarify something in my head, in statics, strictly speaking, do I not consider the moment acting on a point EVERYTIME because I never had to cut the body and the different types of supports (connections) have their unique properties? I'm sorry if that made zero sense.... but I do understand your answer to my question clearly. $\endgroup$ – Skipher Aug 28 '15 at 0:22
  • $\begingroup$ Yes. If you fail to consider bending moment at the cut and take the SUM of bending moments about the point of the cut, you will see that it does not sum to zero (i.e. it will not be in static equilibrium). As you do more free body diagrams and problems you will come to realize that bending moments counter rotation and create equilibrium. I'm not sure about the previous problems you did in statics but is it possible you weren't taking cuts? Maybe you were just doing force balances of free body diagrams (only considering reaction forces and external loads and NOT considering internal forces). $\endgroup$ – pauloz1890 Aug 28 '15 at 0:31
  • $\begingroup$ I think you're getting confused between a standard free body diagram and then taking a cut. Considering internal forces and taking a cut is what seems new to you. Before you were probably just doing a free body diagram to solve reaction forces, am I right? You never considered internal forces and for this reason you never had to add a moment vector to your diagram. @Skipher $\endgroup$ – pauloz1890 Aug 28 '15 at 0:38
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    $\begingroup$ You are correct. I took statics last semester; therefore, mechanics of materials, including the concept of cutting a body to consider the internal forces, is new to me. This is how I now understand why I must consider the moment at the cut area: the moment is being applied by the other cut part on the cut part I'm looking at to resist bending. Because the other cut part applies a moment on this cut part, it is considered an internal force. Is this correct? $\endgroup$ – Skipher Aug 28 '15 at 0:41
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    $\begingroup$ Yes it is true when you take a cut, each side has equal and opposite values of V, N and M. @Skipher This video is not so bad at explaining it if you have the time: youtube.com/watch?v=Z-jQyeYApU8 $\endgroup$ – pauloz1890 Aug 28 '15 at 0:45
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It's useful to be clear about the fact that the 'cut' is an imaginary one. Since the portion of the beam no longer attached to the wall (because of our imaginary cut) is not falling or rotating, it must be subjected to the same type of forces/torques as if it were attached directly to its own 'wall'. The 'wall' in this case is actually the part of the beam on the other side of the imaginary cut though.

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  • $\begingroup$ That's certainly another way to look at it. The latter part of your explanation is actually how I'm understanding this concept. The other part of the beam applied these "reaction forces and moments" like it is a support although it simply has an internal relationship with its counterpart portion. $\endgroup$ – Skipher Aug 28 '15 at 2:24

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